7

Let $S^1$ be the unit sphere $x_1^2+x_2^2=1$ in $\mathbb{R}^2$ and let $X=S^1\times S^1\in\mathbb{R}^4$ with defining equations $f_1=x_1^2+x_2^2-1=0, f_2=x_3^2+x_4^2-1=0$. The vector field $$w=x_1\frac\partial{\partial x_2}-x_2\frac\partial{\partial x_1}+\lambda\left(x_4\frac\partial{\partial x_3}-x_3\frac\partial{\partial x_4}\right)$$ ($\lambda\in\mathbb{R}$) is tangent to $X$ and hence defines by restriction a vector field $v$ on $X$. What is the one-parameter group of diffeomorphisms that $v$ generates?

The definition of a one-parameter group of diffeomorphisms that I'm using is the following:

Let $U$ be an open subset of $\mathbb{R}^n$ and $F : U \times \mathbb{R} \rightarrow U$ a $C^{\infty}$ mapping. The family of mappings $f_t: U \rightarrow U$ , $f_t(x) = F(x, t)$ is said to be a one-parameter group of diffeomorphisms of $U$ if $f_0$ is the identity map and $f_s \cdot f_t = f_{s+t}$ for all s and t.

First of all, I'm confused how this definition can be applied to our situation. The vector field $v$ is not present anywhere in the definition of a one-parameter group of diffeomorphisms. But it has to be relevant somewhere, right?

Community
  • 1
PJ Miller
  • 8,193
  • It is relevant in the way that solution of ODE with vector field $v$ defines a one-parameter group of diffeomorphisms. – Evgeny Nov 11 '13 at 06:42
  • Also, if you consider vector field $x_1 \frac{\partial}{\partial x_2} - x_2 \frac{\partial}{\partial x_1}$ on a plane, it's a linear system with center steady point at origin. One-parameter group of diffeomorphisms in this case describes rotation of points. There is a guess that vector field from your task can be decoupled in rotations along its generating circles :) I bet someone will write that rigourously earlier than me – Evgeny Nov 11 '13 at 07:00
  • Could you please tell, which book is this question belongs to? – Sachchidanand Prasad Jun 10 '18 at 17:28

1 Answers1

4

This vector field corresponds to following ODE in $\mathbb{R}^4$:

$$ \left ( \begin{array}{right} \\ \dot{x_1}\\ \dot{x_2} \\ \dot{x_3} \\ \dot{x_4} \end{array} \right) = \left ( \begin{array}{right} \\ -x_2 \\ x_1 \\ \lambda x_4 \\ -\lambda x_3 \end{array} \right ). $$ Actually, dynamics of $x_1$, $x_2$ is decoupled from dynamics of $x_3$, $x_4$ and equations can be easily solved. If we represent pairs $(x_1, x_2)$ and $(x_3, x_4)$ as complex numbers $z_1 = x_1 + ix_2$ and $z_2 = x_3 + i x_4$ from unit circle in $\mathbb{C}$, then solution can be written in form $(e^{i(t + \phi_1)}, e^{i(-\lambda t + \phi_2)})$, where $\phi_1$ and $\phi_2$ can be derived from initial value problem for ODE. Well, this is it: the one-parameter group of diffeomorphisms generated by this vector field. You take an arbitratry point $x^{(0)} = (x^{(0)}_1, x^{(0)}_2, x^{(0)}_3, x^{(0)}_4)$; it corresponds to $(z^{(0)}_1, z^{(0)}_2)$, where $z^{(0)}_1 = e^{i\phi^{(0)}_1}$, $z^{(0)}_2 = e^{i\phi^{(0)}_2}$. The action $F(x^{(0)}, t)$ is following: $$ F(x^{(0)}, t) = \left ( \mathrm{Re} \left ( e^{i\left(t + \phi^{(0)}_1\right)} \right), \mathrm{Im} \left ( e^{i\left(t + \phi^{(0)}_1\right)} \right), \mathrm{Re} \left ( e^{i\left(-\lambda t + \phi^{(0)}_2\right)}\right), \mathrm{Im}\left ( e^{i\left(-\lambda t + \phi^{(0)}_2\right)} \right) \right). $$ Properties of complex exponenti is a key fact to show that $F(x, t)$ is a one-parameter group of diffeomorphisms.

Evgeny
  • 5,755